A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 3/2 cm and its depth is 8/9 cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.
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Answer:
Volume of metal B=
3
1
πr
1
2
h
1
=
3
1
π(
2
3
)
2
(
9
8
)=
3
2π
Volume of metal A=πr
2
h−
3
1
πr
1
2
h
1
=
3
π
(3r
2
h−r
1
2
h
1
)
=
7×3
22
(3×9×5−
4
9
×89)=
21
22
(135−2)=
21
22×133
=139.33 cm
3
ratio =
2π
139.33×3
=
3.14
209
=
π
209
=
22
209×7
=
2
133
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